NIMCET 2016 Mathematics PYQ — Two common tangents to the circle and parabola are… | Mathem Solvex | Mathem Solvex
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NIMCET 2016 — Mathematics PYQ
NIMCET | Mathematics | 2016
Two common tangents to the circle x2+y2=2a2 and parabola y2=8ax are
Choose the correct answer:
A.
x = ±(y + 2a)
B.
y = ±(x + 2a)
(Correct Answer)
C.
x = ±(y + a)
D.
y = ±(x + a)
Correct Answer:
y = ±(x + 2a)
Explanation
Concept: Standard equation of Parabola y2=4ax Equation of a tangent to the Parabola y = mx + ma, Where m is the slope of the tangent
Equation of a circle with center (0,0) and radius r is x2+y2=r2
Perpendicular distance from centre of a circle to the tangent to a circle is the radius of a circle
Perpendicular distance of a point (x1,y1) from a line Ax+By+C=0 is A2+B2∣Ax1+By1+C∣
Calculation:
Given Equation of parabola y² = 8ax Equation of circle x² + y² = 2a² hence radius r = √2a comparing the Parabola equation with the standard equation
so the equation of a tangent to given parabola is y = mx + m2a ... (1)
Perpendicular distance of the above line from center of circle (0,0) is equal to radius of circle √2a m2a = 2a = 2a√(m² + 1) ⇒ m2a = 2a√(m² + 1)
Squaring both sides and arranging the equation m⁴ + m² - 2 = 0 ⇒ (m² + 2)(m² - 1) = 0 (m² + 2) = 0 not possible as slope cannot be a complex number (m² - 1) = 0 ⇒ m = ±1
put the value of m in equation 1 to get the equation of the common tangent y = ±(x + 2a)
Explanation
Concept: Standard equation of Parabola y2=4ax Equation of a tangent to the Parabola y = mx + ma, Where m is the slope of the tangent
Equation of a circle with center (0,0) and radius r is x2+y2=r2
Perpendicular distance from centre of a circle to the tangent to a circle is the radius of a circle
Perpendicular distance of a point (x1,y1) from a line Ax+By+C=0 is A2+B2∣Ax1+By1+C∣
Calculation:
Given Equation of parabola y² = 8ax Equation of circle x² + y² = 2a² hence radius r = √2a comparing the Parabola equation with the standard equation
so the equation of a tangent to given parabola is y = mx + m2a ... (1)
Perpendicular distance of the above line from center of circle (0,0) is equal to radius of circle √2a m2a = 2a = 2a√(m² + 1) ⇒ m2a = 2a√(m² + 1)
Squaring both sides and arranging the equation m⁴ + m² - 2 = 0 ⇒ (m² + 2)(m² - 1) = 0 (m² + 2) = 0 not possible as slope cannot be a complex number (m² - 1) = 0 ⇒ m = ±1
put the value of m in equation 1 to get the equation of the common tangent y = ±(x + 2a)
